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ABCD is a square. E and F are respective...

ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF, prove that `ar (DeltaAER) = ar (DeltaAFR)`.

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In `Delta ABE and Delta ADF`,
`{:( :'),( :'),( :'):}{(AB=AD,"(sides of a square are equal)"),(angleABE = angleADF,("each" 90^(@))),(" "BE = DF,):}`
( `:'` E is the mid-point of BC and F is the mid-poiunt of CD. Also, `(1)/(2) BC = (1)/(2) CD`)
`:. Delta ABE ~= Delta ADF` (by SAS congruence rule)
`:. AE = AF` (c.p.c.t)...(1)
Now, in `Delta AER and Delta AFR`,
`:'{(AE = AF," [from(1)]"),(ER = RF,( :' "R is mid-point of EF")),(AR = AR,"(common side)"):}`
`:. Delta AER = Delta AFR` (by SSS rule of congruence)
Hence, `ar(Delta AER) = ar(Delta AFR)` ( `:'` congruent triangle are equal in area)
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